2012-11-04 17:40:49Morris
[ITSA] 18th 總解答
一個人比賽，還用得著去寫旋轉就太瘋了，果斷打表。
#include <stdio.h>
char num[10][6][6] = {
{"01110", "10001", "10001", "10001", "01110"},//0
{"00100", "00100", "00100", "00100", "00100"},//1
{"01110", "10001", "00010", "00100", "11111"},//2
{"01110", "10001", "00110", "10001", "01110"},//3
{"10010", "10010", "01111", "00010", "00010"},//4
{"11111", "10000", "11110", "00001", "11110"},//5
{"00010", "00100", "01110", "10001", "01110"},//6
{"01110", "10001", "00010", "00010", "00010"},//7
{"01110", "10001", "01110", "10001", "01110"},//8
{"01110", "10001", "01110", "00010", "00010"},//9
};
int main() {
int t;
scanf("%d", &t);
while(t--) {
int i, j;
int a, b, c, x;
int g[5][5];
for(i = 0; i < 5; i++) {
for(j = 0; j < 5; j++) {
scanf("%d", &x);
g[i][j] = x+'0';
}
}
int tmp, ans = 0;
for(i = 0; i < 30; i++) {
for(j = 0; j < 10; j++) {
for(a = 0; a < 10; a++) {
int cnt = 0;
for(b = 0; b < 5; b++)
for(c = 0; c < 5; c++)
if(g[b][c] == num[a][b][c])
cnt++;
if(cnt == 25)
ans = a;
}
tmp = g[1][1];
g[1][1] = g[1][2];
g[1][2] = g[1][3];
g[1][3] = g[2][3];
g[2][3] = g[3][3];
g[3][3] = g[3][2];
g[3][2] = g[3][1];
g[3][1] = g[2][1];
g[2][1] = tmp;
}
tmp = g[0][0];
g[0][0] = g[0][1];
g[0][0] = g[0][1];
g[0][1] = g[0][2];
g[0][2] = g[0][3];
g[0][3] = g[0][4];
g[0][4] = g[1][4];
g[1][4] = g[2][4];
g[2][4] = g[3][4];
g[3][4] = g[4][4];
g[4][4] = g[4][3];
g[4][3] = g[4][2];
g[4][2] = g[4][1];
g[4][1] = g[4][0];
g[4][0] = g[3][0];
g[3][0] = g[2][0];
g[2][0] = g[1][0];
g[1][0] = tmp;
}
printf("%d\n", ans);
}
return 0;
}
樹形 dp，事實上這題在比賽的時候，測資是錯誤的，用錯誤的方法也會過，根據通過的隊伍獲得資料，測資筆數只有一筆，而且 n 還小於 10，由於主辦單位給了多餘的輸入，以至於多輸出了答案，無語現場摔鍵盤。
很多人都說測資錯了，主辦單位還不信，只好認命點了，這次沒破台。
#include <stdio.h>
long long dp1[90005], dp2[90005];
int p[90005], son[90005];
int main() {
int n, i;
while(scanf("%d", &n) == 1) {
for(i = 1; i < n; i++) {
scanf("%d", &p[i]);
dp1[i] = dp2[i] = son[i] = 0;
}
dp1[0] = dp2[0] = son[0] = 0;
for(i = n-1; i > 0; i--) {
son[i]++;
dp1[p[i]] += dp1[i] + son[i];
son[p[i]] += son[i];
}
for(i = 1; i < n; i++) {
dp2[i] = (dp2[p[i]]+n-son[i]) + (dp1[p[i]]-dp1[i]-son[i]);
}
long long ans = dp1[0]+dp2[0];
int lab = 0;
for(i = 1; i < n; i++) {
if(dp1[i]+dp2[i] < ans)
ans = dp1[i]+dp2[i], lab = i;
}
printf("%d\n", lab);
}
return 0;
}
這題大家肯定是水了他。
#include <stdio.h>
int main() {
char s[1000];
int i;
while(gets(s)) {
int cnt[26] = {};
for(i = 0; s[i]; i++) {
if(s[i] >= 'A' && s[i] <= 'Z')
cnt[s[i]-'A']++;
else
if(s[i] >= 'a' && s[i] <= 'z')
cnt[s[i]-'a']++;
}
for(i = 0; i < 26; i++) {
if(cnt[i])
printf("%c %d\n", i+'a', cnt[i]);
}
}
return 0;
}
其實這題答案不會超過 n，證明可以簡單地得到，而且好像用 greedy 就可以知道答案為多少，但是我居然用 dfs 跑，而且還能通過，我就沒多說什麼了。
#include <stdio.h>
int A[100], i, n;
int ans;
void dfs(int dep) {
if(dep >= ans)
return;
int flag = 0, i, tmp, j;
for(i = 0; i < n; i++) {
if(A[i] != i) {
j = A[i];
tmp = A[i], A[i] = A[j], A[j] = tmp;
dfs(dep+1);
flag = 1;
tmp = A[i], A[i] = A[j], A[j] = tmp;
}
}
if(flag == 0) {
if(ans > dep)
ans = dep;
}
}
int main() {
while(scanf("%d", &n) == 1) {
ans = n*n;
for(i = 0; i < n; i++) {
scanf("%d", &A[i]);
}
dfs(0);
printf("%d\n", ans);
}
return 0;
}
這一題是你猜我猜，連工作人員都在猜，在看不懂題目的意思下，猜一猜賭賭看囉。
#include <stdio.h>
int main() {
int n;
while(scanf("%d", &n) == 1) {
int deg[90005] = {}, x, y, i;
for(i = 1; i < n; i++) {
scanf("%d %d", &x, &y);
deg[x]++, deg[y]++;
}
int ans = 0;
for(i = 0; i < n; i++) {
if(deg[i] >= 2)
ans++;
}
printf("%d\n", ans);
}
return 0;
}
#include <stdio.h>
char num[10][6][6] = {
{"01110", "10001", "10001", "10001", "01110"},//0
{"00100", "00100", "00100", "00100", "00100"},//1
{"01110", "10001", "00010", "00100", "11111"},//2
{"01110", "10001", "00110", "10001", "01110"},//3
{"10010", "10010", "01111", "00010", "00010"},//4
{"11111", "10000", "11110", "00001", "11110"},//5
{"00010", "00100", "01110", "10001", "01110"},//6
{"01110", "10001", "00010", "00010", "00010"},//7
{"01110", "10001", "01110", "10001", "01110"},//8
{"01110", "10001", "01110", "00010", "00010"},//9
};
int main() {
int t;
scanf("%d", &t);
while(t--) {
int i, j;
int a, b, c, x;
int g[5][5];
for(i = 0; i < 5; i++) {
for(j = 0; j < 5; j++) {
scanf("%d", &x);
g[i][j] = x+'0';
}
}
int tmp, ans = 0;
for(i = 0; i < 30; i++) {
for(j = 0; j < 10; j++) {
for(a = 0; a < 10; a++) {
int cnt = 0;
for(b = 0; b < 5; b++)
for(c = 0; c < 5; c++)
if(g[b][c] == num[a][b][c])
cnt++;
if(cnt == 25)
ans = a;
}
tmp = g[1][1];
g[1][1] = g[1][2];
g[1][2] = g[1][3];
g[1][3] = g[2][3];
g[2][3] = g[3][3];
g[3][3] = g[3][2];
g[3][2] = g[3][1];
g[3][1] = g[2][1];
g[2][1] = tmp;
}
tmp = g[0][0];
g[0][0] = g[0][1];
g[0][0] = g[0][1];
g[0][1] = g[0][2];
g[0][2] = g[0][3];
g[0][3] = g[0][4];
g[0][4] = g[1][4];
g[1][4] = g[2][4];
g[2][4] = g[3][4];
g[3][4] = g[4][4];
g[4][4] = g[4][3];
g[4][3] = g[4][2];
g[4][2] = g[4][1];
g[4][1] = g[4][0];
g[4][0] = g[3][0];
g[3][0] = g[2][0];
g[2][0] = g[1][0];
g[1][0] = tmp;
}
printf("%d\n", ans);
}
return 0;
}
樹形 dp，事實上這題在比賽的時候，測資是錯誤的，用錯誤的方法也會過，根據通過的隊伍獲得資料，測資筆數只有一筆，而且 n 還小於 10，由於主辦單位給了多餘的輸入，以至於多輸出了答案，無語現場摔鍵盤。
很多人都說測資錯了，主辦單位還不信，只好認命點了，這次沒破台。
#include <stdio.h>
long long dp1[90005], dp2[90005];
int p[90005], son[90005];
int main() {
int n, i;
while(scanf("%d", &n) == 1) {
for(i = 1; i < n; i++) {
scanf("%d", &p[i]);
dp1[i] = dp2[i] = son[i] = 0;
}
dp1[0] = dp2[0] = son[0] = 0;
for(i = n-1; i > 0; i--) {
son[i]++;
dp1[p[i]] += dp1[i] + son[i];
son[p[i]] += son[i];
}
for(i = 1; i < n; i++) {
dp2[i] = (dp2[p[i]]+n-son[i]) + (dp1[p[i]]-dp1[i]-son[i]);
}
long long ans = dp1[0]+dp2[0];
int lab = 0;
for(i = 1; i < n; i++) {
if(dp1[i]+dp2[i] < ans)
ans = dp1[i]+dp2[i], lab = i;
}
printf("%d\n", lab);
}
return 0;
}
這題大家肯定是水了他。
#include <stdio.h>
int main() {
char s[1000];
int i;
while(gets(s)) {
int cnt[26] = {};
for(i = 0; s[i]; i++) {
if(s[i] >= 'A' && s[i] <= 'Z')
cnt[s[i]-'A']++;
else
if(s[i] >= 'a' && s[i] <= 'z')
cnt[s[i]-'a']++;
}
for(i = 0; i < 26; i++) {
if(cnt[i])
printf("%c %d\n", i+'a', cnt[i]);
}
}
return 0;
}
其實這題答案不會超過 n，證明可以簡單地得到，而且好像用 greedy 就可以知道答案為多少，但是我居然用 dfs 跑，而且還能通過，我就沒多說什麼了。
#include <stdio.h>
int A[100], i, n;
int ans;
void dfs(int dep) {
if(dep >= ans)
return;
int flag = 0, i, tmp, j;
for(i = 0; i < n; i++) {
if(A[i] != i) {
j = A[i];
tmp = A[i], A[i] = A[j], A[j] = tmp;
dfs(dep+1);
flag = 1;
tmp = A[i], A[i] = A[j], A[j] = tmp;
}
}
if(flag == 0) {
if(ans > dep)
ans = dep;
}
}
int main() {
while(scanf("%d", &n) == 1) {
ans = n*n;
for(i = 0; i < n; i++) {
scanf("%d", &A[i]);
}
dfs(0);
printf("%d\n", ans);
}
return 0;
}
這一題是你猜我猜，連工作人員都在猜，在看不懂題目的意思下，猜一猜賭賭看囉。
#include <stdio.h>
int main() {
int n;
while(scanf("%d", &n) == 1) {
int deg[90005] = {}, x, y, i;
for(i = 1; i < n; i++) {
scanf("%d %d", &x, &y);
deg[x]++, deg[y]++;
}
int ans = 0;
for(i = 0; i < n; i++) {
if(deg[i] >= 2)
ans++;
}
printf("%d\n", ans);
}
return 0;
}